# Simultaneous equations.

1. Sep 22, 2009

### camino

1. The problem statement, all variables and given/known data

160 cos(theta) - W cos(69) = 0

160 sin(theta) - W cos(21) - W = 0

Can someone solve these equations for me to find W and theta?

2. Sep 22, 2009

### rock.freak667

160cosθ=Wcos(69)

W=160cosθ/cos(69)

continue now

3. Sep 22, 2009

### camino

W = W cos(69) / cos(69)

?

4. Sep 22, 2009

### rock.freak667

no, put W=160cosθ/cos(69) into the second equation

5. Sep 22, 2009

### camino

Sorry I am still not seeing it. Could you give me one more step?

6. Sep 22, 2009

### rock.freak667

In this equation

160 sinθ - W cos(21) - W = 0

everywhere you see 'W' put 160cosθ/cos(69) and try to simplify it.

7. Sep 22, 2009

### camino

I do understand how to simplify one equation and substitute into the other, however I just can't seem to simplify these down. The theta with the sin and cos attached is what is really confusing me when trying to simplify. This is just a mess and I'm very confused.

8. Sep 22, 2009

### rock.freak667

just show where you've reached and we'll see if we can simplify it further

9. Sep 22, 2009

### camino

160 sinθ - (160cosθ/cos69) cos(21) - (160cosθ/cos69) = 0

10. Sep 22, 2009

### rock.freak667

so you get

$$160 sin \theta - (\frac{160cos21}{cos69} - \frac{160}{cos69})cos \theta = 0$$

this is like Asinθ-Bcosθ=0, how do you solve this equation?

11. Sep 22, 2009

### camino

Is there a trig identity that replaces either sin or cos? I'm thinking we want to narrow this down to only having either sin or cos in the equation.

12. Sep 22, 2009

### rock.freak667

well yes but we can put it like this

Asinθ=Bcosθ

or sinθ/cosθ=B/A

do we know another way to express sinθ/cosθ ?

13. Sep 22, 2009

### camino

I am really not sure.

14. Sep 22, 2009

### rock.freak667

do you not know that tanθ=sinθ/cosθ ?

15. Sep 22, 2009

### camino

Oh my it has been a long day.. Yes i see that now. So simplifying I would then do:

160 sinθ - (-29.6542) cosθ = 0

160 sinθ = 29.6542 cosθ

θ = tan^-1(160/29.6542)

θ = 79.5° ?

16. Sep 22, 2009

### rock.freak667

yes that should be correct. You are able to find W knowing θ=79.5° right?

17. Sep 22, 2009

### camino

Yes:

160cos(79.5) - W cos(69) = 0

29.1577 = W cos(69)

W = 81.36

18. Sep 22, 2009

### camino

Thank you so much for all your help!